Q76 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { If } \sin A=\dfrac{4}{5} \text { and } \cos B=\dfrac{-12}{13} \text { where } A \text { and } B \text { lie in first and third quadrant respectively. Then } \cos (A+B)=$
- Correct answerA$\dfrac{-16}{65}$
- B$\dfrac{-56}{65}$
- C$\dfrac{16}{65}$
- D$\dfrac{56}{65}$
Solution
Given $\sin A = \dfrac{4}{5}$ and $A$ is in the first quadrant, $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \sqrt{1 - \dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$.
Given $\cos B = \dfrac{-12}{13}$ and $B$ is in the third quadrant, $\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \left(\dfrac{-12}{13}\right)^2} = -\sqrt{1 - \dfrac{144}{169}} = -\sqrt{\dfrac{25}{169}} = -\dfrac{5}{13}$.
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$, we substitute the values:
$\cos(A + B) = \left(\dfrac{3}{5}\right) \left(\dfrac{-12}{13}\right) - \left(\dfrac{4}{5}\right) \left(\dfrac{-5}{13}\right)$
$\cos(A + B) = \dfrac{-36}{65} - \left(\dfrac{-20}{65}\right)$
$\cos(A + B) = \dfrac{-36 + 20}{65} = \dfrac{-16}{65}$.
Answer: $\dfrac{-16}{65}$
Given $\cos B = \dfrac{-12}{13}$ and $B$ is in the third quadrant, $\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \left(\dfrac{-12}{13}\right)^2} = -\sqrt{1 - \dfrac{144}{169}} = -\sqrt{\dfrac{25}{169}} = -\dfrac{5}{13}$.
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$, we substitute the values:
$\cos(A + B) = \left(\dfrac{3}{5}\right) \left(\dfrac{-12}{13}\right) - \left(\dfrac{4}{5}\right) \left(\dfrac{-5}{13}\right)$
$\cos(A + B) = \dfrac{-36}{65} - \left(\dfrac{-20}{65}\right)$
$\cos(A + B) = \dfrac{-36 + 20}{65} = \dfrac{-16}{65}$.
Answer: $\dfrac{-16}{65}$
Hence, the correct answer is option: 'A'.